Applied Statistics Probability Theory MPS Department | FEU Institute of Technology Applied Statistics Basic Concepts MPS Department | FEU Institute of Technology OBJECTIVES ▪ ▪ ▪ ▪ Define elements of an experiment Identify elements of a Sample Space Differentiate events and Sample Space Perform operations on events of an Experiment An experiment is the process of observing a phenomenon that has variation in its outcomes. It is a well-defined action leading to a single, well-defined result. An outcome is a result from a single trial of an experiment The set of all possible outcomes of an experiment is called the Sample Space, represented by S An event is a collection of some outcome from an experiment. It is a subset of the sample space. An event containing only one element is called a simple event. A compound event is one that can be expressed as a union of simple events. The null space or empty space is a subset of the sample space that contains no elements. It is denoted by ∅ The union of two events A and B, denoted by 𝐴 ∪ 𝐵, is the event containing all elements that belong to both A or to B, or to both The complement of an event A is the set of all elements of S that are not in A. We denote this by A’ The intersection of two events A and B, denoted by 𝐴 ∩ 𝐵 , is the event containing all elements common to both A and B Two events A and B are mutually exclusive if 𝐴 ∩ 𝐵 = ∅; that is, A and B have no elements in common List the elements of the following sample spaces: a. The set of integers between 1 and 50 divisible by 8 S={8, 16, 24, 32, 40, 48} b. The Set of outcomes when a coin is tossed until a tail or three heads appear. S={T, HT, HHT, HHH} A newly married couple is planning to have three children. List the elements of the sample space S using M for ‘male’ and F for ‘female’. Define a second sample space X, where the elements represent the number of females S={MMM, MMF, MFM, FMM, MFF, FMF, FFM, FFF} X={ 0, 1, 2, 3} In an experiment involving rolling a single die, let X be the event of getting an even number of dots facing up, and Y be the event of getting less than 4 dots facing up. What is 𝑋 ∪ 𝑌? 𝑋 ∩ 𝑌? And X’? S={1,2,3,4,5,6} X={2,4,6} Y={1,2,3} This means 𝑋 ∪ 𝑌 = 1,2,3,4,6 𝑋 ∩ 𝑌 = {2} 𝑋 ′ = {1,3,5} In a dice-roll experiment, let E be the event when the number of dots facing up is even, M be the event when the number of dots is less than 3. Then 𝑆 = {1,2,3,4,5,6}; 𝐸 = {2,4,6}; 𝑀 = {1,2,3} Now 𝐸’ = {1,3,5}; 𝑀’ = {4,5,6}; 𝐸 ∪ 𝑀 = 1,2,3,6 ; 𝐸 ∩ 𝑀 = {2} Applied Statistics Counting Principles MPS Department | FEU Institute of Technology OBJECTIVES ▪ ▪ ▪ Discuss the basic concepts on counting Differentiate and illustrate permutations and combinations Apply theorems and formulas on counting elements of the Sample Space in a given Experiment If an operation can be performed in 𝑛1 ways, and if for each of these, a second operation can be performed in 𝑛2 ways, then the two operations can be performed together in 𝑛1 𝑛2 ways In a medical study, patients are classified in 8 ways according to blood-type - 𝐴𝐵 + , 𝐴𝐵 − , 𝐴+ , 𝐴− , 𝐵 + , 𝐵− , 𝑂+ , 𝑂− ; and also according to whether their blood pressure is low, normal, or high. Find the number of ways in which a patient can be classified Solution: Classifying a patient according to blood type occurs in 8 ways, and according to blood pressure in 3 ways. By multiplication rule, there are 8 ∙ 3 = 24 ways to classify a patient If an experiment consists of throwing a die and then drawing a letter at random from the English alphabet. How many outcomes are in the Sample Space? Solution: There are 6 results in throwing a die, and 26 possible letter on the English alphabet. Hence, there are 6 ∙ 26 = 156 elements in the sample space If an operation can be performed in 𝑛1 ways. If for each of these, a second operation can be performed in 𝑛2 ways. If for each of the first two a third operation can be performed in 𝑛3 ways, and so forth. Then a sequence of 𝑘 operations can be performed in 𝑛1 𝑛2 𝑛3 … 𝑛𝑘 ways. A college freshman must take a Science course, a Humanities course, and a Mathematics course. If he may select any of 6 Science courses, any of 4 Humanities, and any of 4 Mathematics courses, how many ways can he arrange his program? Solution: There are 6 choices for a Science course, 4 for Humanities and 4 for Math. By the Fundamental Countering Principle, there are 6 ∙ 4 ∙ 4 = 96 possible arrangements. In how many ways can a true-or-false test consisting of 10 questions be answered? Solution: For each question, there are 2 possible answers, hence, there are 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 = 210 = 1024 possibilities A Permutation is an ordered arrangement of all or part of a set of objects. Theorem 1: The number of permutations of 𝑛 distinct objects is 𝑛! = 𝑛 𝑛 − 1 𝑛 − 2 … ∙ 3 ∙ 2 ∙ 1 How many distinct permutations can be made from the letters of the word ‘COLUMNS’? There are 7 objects to be arranged. Hence there are 7! = 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 5040 possible permutations How many of these permutations start with ‘M’? Fixing the first letter as M means we randomly arrange the 6 remaining letters. Hence there are 6! = 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 720 of those that start with M Theorem 2 The number of permutations of 𝑛 objects taken 𝑟 at a time is 𝑛! 𝑛𝑷𝑟 = 𝑛−𝑟 ! Find the number of ways in which 6 teachers can be assigned to 4 sections of an introductory psychology course if no teacher is assigned to more than one section? Solution: Since we have 6 options but takes only 4 of them, there are 6𝑃4 = 360 possible assignments Three lottery tickets are drawn from 40 for first, second and third prizes. Find the number of sample points in S for awarding the three prizes if no contestant can win more than one prize? Answer: 40P3=59,280 Theorem 3: The number of permutations of 𝑛 distinct objects arranged in a circle is is (𝑛 − 1)! In how many ways can a caravan of 8 covered wagons from Arizona be arranged in a circle? (8-1)!=7!=5040 How many ways can 5 different trees be planted in a circle? (5-1)!=4!=24 Theorem 4 The number of distinct permutations of 𝑛 things of which 𝑛1 are of one kind, 𝑛2 of a second kind,…, 𝑛𝑘 of a kth kind is 𝑛! 𝑛1 ! 𝑛2 ! … 𝑛𝑘 ! How many distinct permutations can be made from the letters of the word ‘statistics’? Solution: There are 10 letters, 3 of which are ‘s’; 3 are ‘t’; 2 are ‘i’ one c and one a. hence, there are 10! = 50400 3! 3! 2! 1! 1! distinct permutations In how many ways can 3 oaks, 4 pines, and 2 maples be arranged along a property line if one does not distinguish between trees of the same kind? Solution: 9! = 1260 3! 4! 2! possible arrangements Theorem 5 The number of ways of partitioning a set of 𝑛 objects into 𝑟 cells with 𝑛1 elements on the first cell, 𝑛2 elements on the second, and so on is 𝑛! where 𝑛1 + 𝑛2 + ⋯ + 𝑛𝑟 = 𝑛 𝑛1 ! 𝑛2 ! … 𝑛𝑟 ! A college plays 12 football games during a season. In how many ways can the team end the season with 7 wins, 3 loses and 2 ties? Solution: 12! = 7920 7! 3! 2! possible ways Nine people are going on a picnic trip in three cars that can hold 2, 4, and 3 passengers respectively. How many ways is it possible to transport the 9 people to the picnic site using all cars? Solution: 9! = 1260 2! 4! 3! possible ways A combination is an arrangement of objects without regard to order. Theorem 6 The number of combinations of 𝑛 distinct objects taken 𝑟 at a time is 𝑛! 𝑛𝑪𝑟 = 𝑛 − 𝑟 ! 𝑟! How many ways are there to select 3 candidates from 8 equally qualified recent graduates for openings in a firm? Answer: 8𝐶3 = 56 From a group of 5 men and 3 women, how many committees of size 3 are possible with a. With no restrictions? 8𝐶3 = 56 b. with 1 man and 2 women? 5𝐶1 3𝐶2 = 15 Applied Statistics Probability of an Event MPS Department | FEU Institute of Technology OBJECTIVES ▪ ▪ Differentiate multiplication rule from addition rule of probability Illustrate dependent, independent and mutually exclusive events and compute their probabilities Probability refers to the likelihood of occurrence of an event. There are three approaches to Probability 1. Subjective Probability – chance of occurrence is given by a particular person based on his/her educated guess, opinion, intuition or beliefs. 2. Empirical Probability – probability is assigned based on the prior knowledge of the events that happened on the past, or based on research or experiment 3. Classical Probability – applied when all possible outcomes are equally likely to happen. In Classical Probability, the Probability that an event E will occur is 𝑛(𝐸) number of outcomes in E 𝑃 𝐸 = = 𝑛(𝑆) number of outcomes in the Sample Space The following are properties of Probability: (i) 0 ≤ 𝑃 𝐸 ≤ 1, 𝑃 ∅ = 0, 𝑃 𝑆 = 1 (ii) If 𝑆 = {𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑘 }, then 𝑃 𝑥𝑖 = 1/𝑘 and σ 𝑃 𝑥𝑖 = 𝑃 𝑥1 + 𝑃 𝑥2 + ⋯ + 𝑃 𝑥𝑖 = 1 Theorem 7 If E and E’ are complementary events, then 𝑃 𝐸 + 𝑃 𝐸 ′ = 1, 𝑜𝑟 𝑃 𝐸 ′ = 1 − 𝑃(𝐸) Find the errors in each of the following statements: 1. The probabilities that an automobile salesperson will sell 1,2,3, or 4 cars on any given day are, respectively, 0.19, 0.387, 0.29, and 0.15 0.19+0.387+0.29+0.15=1.015 2. The probability that it will rain today is 0.40, and the probability that it will not rain is 0.52 0.52+0.40 = 0.92 Choose a number at random from 1 to 13. What is the probability that a. The number is even? b. The number is less than 5? c. The number is greater than 6? Solution: The experiment can result in 13 equally likely outcomes. a. Since there are 6 even numbers, then P(even)=6/13 b. Four numbers are less than 5, so P(< 5)=4/13 c. Seven numbers are greater than 6, hence P(>6)=7/13 If a letter is chosen at random from the English alphabet, What is the probability that the letter a. is a vowel? b. precedes the letter ‘J’? c. follows the letter ‘G’? Solution: There are 26 letters in the English alphabet a. 5 of them are vowels, hence P(vowel)=5/26 b. 9 preceeds ‘J’, hence P(preceeds J)=9/26 c. 19 follows ‘G’, hence P(follows G)=19/26 A die is loaded in such a way that an even number is twice as likely to occur as an odd number. If E is an event that a number is less than 4 occurs on a single toss, find P(E) Solution: Let P(odd)=w, then P(even)=2w. This gives P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=1 w + 2w + w + 2w + w + 2w =1 9w=1 means w=1/9. Hence P(odd)=1/9 and P(even)=2/9 Now P(E)=P(1)+P(2)+P(3) = 1/9 + 2/9+ 1/9 = 4/9 If a card is drawn from an ordinary deck, What is the probability of drawing a. An ace? b. A spade? c. A face card? 4/52= 1/13 13/52= 1/4 12/52=3/13 If three books are picked at random from a shelf containing 5 novels, 3 book of poems, 2 dictionaries, what is the probability that A novel, a book of poem, and a dictionary are selected? Solution: The number of ways on Selecting 3 books from a shelf with 10 is 10C3. The number of ways of selecting a novel, a poem and a dictionary on that shelf is (5C1)(3C1)(2C1). Hence, the required probability is (5𝐶1)(3𝐶1)(2𝐶1) 1 𝑃 𝐸 = = 10𝐶3 4 If three books are picked at random from a shelf containing 5 novels, 3 book of poems, 2 dictionaries, what is the probability that The three book of poems are selected? Solution: The number of ways on Selecting 3 books from a shelf with 10 is 10C3. The number of ways of the three poems on that shelf is (3C3). Hence, the required probability is (3𝐶3) 1 𝑃 𝐸 = = 10𝐶3 120 If three books are picked at random from a shelf containing 5 novels, 3 book of poems, 2 dictionaries, what is the probability that 2 novels and a book of poems are selected? Solution: The number of ways on Selecting 3 books from a shelf with 10 is 10C3. The number of ways of selecting 2 novels and a book of poems on that shelf is (5C2)(3C1). Hence, the required probability is (5𝐶2)(3𝐶1) 1 𝑃 𝐸 = = 10𝐶3 4 Theorem 8 If A and B are two events, then 𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 ∩ 𝐵) In a high school computer class, there are 15 juniors and 10 seniors. Four juniors and five seniors are boys. If a student is selected at random, what is the probability of selecting a junior or a boy? Solution: P(Junior)=15/25, P(boy)=9/25, P(Junior ∩ Boy)=4/25. By addition rule, 15 9 4 4 𝑃 𝐽∪ 𝐵 =𝑃 𝐽 +𝑃 𝐵 – 𝑃 𝐽 ∩ 𝐵 = + − = 25 25 25 5 The probability that an American industry will locate in Manila is 0.75, the probability that it will locate in Baguio is 0.38, and the probability that it will locate in either Manila or Baguio is 0.8. What is the probability that the industry will locate in both cities? Solution: Given 𝑃 𝑀 = 0.75, 𝑃 𝐵 = 0.38, 𝑃 𝑀 ∪ 𝐵 = 0.8 By the addition rule, 𝑃 𝑀 ∪ 𝐵 = 𝑃 𝑀 + 𝑃 𝐵 − 𝑃 𝑀 ∩ 𝐵 This gives 0.8 = 0.75 + 0.38 − 𝑃(𝑀 ∩ 𝐵) Hence 𝑃 𝑀 ∩ 𝐵 = 0.75 + 0.38 − 0.8 = 0.33 Corollary8.1: If A and B are mutually exclusive, then 𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃(𝐵) Corollary8.2: If 𝐴1 , 𝐴2 , 𝐴3 , … , 𝐴𝑛 are mutually exclusive events, then 𝑃 𝐴1 ∪ 𝐴2 ∪ ⋯ ∪ 𝐴𝑛 = 𝑃 𝐴1 + 𝑃 𝐴2 + ⋯ + 𝑃(𝐴𝑛 ) If 𝐴 and B are two mutually exclusive events, and 𝑃(𝐴) = 0.7, 𝑃(𝐵) = 0.2, find 𝑃(𝐴 𝑈 𝐵), 𝑃(𝐴’)/ Solution: 𝑃(𝐴 𝑈 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) = 0.7 + 0.2 = 0.9 (Corr 8.1) 𝑃(𝐴’) = 1 − 𝑃(𝐴) = 1 − 0.7 = 0.3 (Rule of Complements) In a college graduating class of 100 students, 54 studied mathematics, 69 studied history, and 35 studied both mathematics and history. If one of these students is selected at random, find the probability that a. The student took mathematics or history b. The student did not take any of these courses c. The student took history but not mathematics The Probability of an event B occurring when it is known that some event A has occurred is called a Conditional Probability that B occurs given A has occurred, denoted by P(B|A) Conditional Probability provides us with a way to reason about the outcome of an experiment, based on partial information. Here are some examples: 1. In an experiment involving two successive rolls of a die, you are told that the sum of the two rolls is 10. How likely is it that the second roll was a 4? 2. In a word guessing game, the first letter of a word is a ‘c’. What is the likelihood that the second letter is an ‘r’? 3. How likely is it that a person has a certain disease given that the medical test was a negative? Let A and B be events, the probability of B, given A, denoted by 𝑃(𝐵|𝐴) is given by 𝑃(𝐴 ∩ 𝐵) 𝑃 𝐵𝐴 = , if 𝑃 𝐴 > 0 𝑃(𝐴) In Metro Manila, 48% of all teenagers own a skateboard and 39% of all teenagers own a skateboard and roller blades. What is the probability that a teenager owns roller blades given that the teenager owns a skateboard? Solution: Let 𝑆 be the event that a teenager owns a skateboard and 𝑅 be the event that a teenager owns rollerblades. Then P(S)=0.48 and 𝑃 𝑆 ∩ 𝑅 = 0.39. The Required Probability is now 𝑃(𝑅 ∩ 𝑆) 0.39 𝑃 𝑅𝑆 = = = 0.8125 𝑃(𝑆) 0.48 Theorem 9 If A and B are two events, then 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 𝑃(𝐵|𝐴) The Probability that the stock market goes up on Tuesday is 0.6. Given that it goes up on Tuesday, the Probability that it goes up on Wednesday is 0.3. Find the probability that it goes up in both days. Solution: Let T=stock goes up on Tue. W=stock goes up on Wed. Given: 𝑃(𝑇) = 0.6; 𝑃(𝑊|𝑇) = 0.3. Now 𝑃(𝑇 ∩ 𝑊) = 𝑃(𝑇)𝑃(𝑊|𝑇) = (0.6)(0.3) = 0.18 Two events A and B are independent if the fact that A occurs does not affect the probability of B occurring, otherwise they are dependent. 1. Rolling a 4 on a single die, and then 1 on the second roll of the die 2. Drawing a face card from a standard deck, replacing it, then drawing an ace as a second card. 1. Picking a red marble from a jar, and without replacement, picking a blue marble for the second. 2. Drawing a face card from an ordinary deck, without replacing, drawing an ace as the second card Theorem 10 Two events A and B are independent if and only if 𝑃 𝐵 𝐴 = 𝑃 𝐵 𝑎𝑛𝑑 𝑃 𝐴 𝐵 = 𝑃 𝐴 . Otherwise, A and B are dependent Theorem 11 If A and B are independent events, then 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 𝑃(𝐵) The probability that Carlo will pass a certain course is 0.90 and that Rina will pass the same course is 0.83. If we assume independence for both events, what is the probability that a. Both will pass the course 𝑃 𝐶 ∩ 𝑅 = 𝑃 𝐶 𝑃 𝑅 = 0.9 0.83 = 0.747 b. Only Carlo will pass the course 𝑃 𝐶 ∩ 𝑅′ = 𝑃 𝐶 𝑃 𝑅′ = 0.9 1 − 0.83 = 0.153 Theorem 12 If, in an experiment, the events 𝐴1 , 𝐴2 , … , 𝐴𝑘 can occur, then 𝑃 𝐴1 ∩ 𝐴2 ∩ ⋯ ∩ 𝐴𝑘 = 𝑃 𝐴1 𝑃 𝐴2 𝐴1 … 𝑃 𝐴𝑘 𝐴1 ∩ 𝐴2 ∩ ⋯ ∩ 𝐴𝑘−1 If the events 𝐴1 , 𝐴2 , … , 𝐴𝑘 are independent, then 𝑃 𝐴1 ∩ 𝐴2 ∩ ⋯ ∩ 𝐴𝑘 = 𝑃 𝐴1 𝑃 𝐴2 … 𝑃(𝐴𝑘 ) Three cards are chosen at random from a deck without replacement. What is the probability of choosing a king, a seven, and an ace, in order? Solution: Let K=a king on the first draw; S=seven on 2nd Draw, A=ace on the 3rd draw. 4 4 4 8 𝑃 𝐾∩𝑆∩𝐴 =𝑃 𝐾 𝑃 𝑆 𝐾 𝑃 𝐴 𝑆∩𝐾 = = 52 51 50 16,575 [1] Bluman, Allan G.; Elementary statistics : a step by step approach, 7th ed, McGraw-Hill 2007 [2] Stephens, Larry J.; Schaum’s outline of theory and problems of beginning statistics McGraw-Hill Companies, Inc. 1998